a: Thay x=49 vào A, ta được:
\(A=\dfrac{\sqrt{49}+1}{\sqrt{49}+3}=\dfrac{7+1}{7+3}=\dfrac{8}{10}=\dfrac{4}{5}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x+2}{x-\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}-x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
c: P=A*B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+3}\)
\(P< =\dfrac{1}{x+3}\)
=>\(\dfrac{1}{\sqrt{x}+3}< =\dfrac{1}{x+3}\)
=>\(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{x+3}< =0\)
=>\(\dfrac{x+3-\sqrt{x}-3}{\left(x+3\right)\left(\sqrt{x}+3\right)}< =0\)
=>\(x-\sqrt{x}< =0\)
=>\(\sqrt{x}\left(\sqrt{x}-1\right)< =0\)
=>\(\sqrt{x}-1< =0\)
=>0<=x<=1
